April 1987Volume 11 Part 2

FU:NCTION

\ A SCHOOL MATHEMATICS MAGAZINEpublished by Monash University

Reg. by Aust Post Publ. No. VBH0111

CONTENTS

The front cover. 34

The square root .of zero.

Clear thinking about limitsa

Simple rational numbers.

Michael A.B»Deakin

Alan Pryde"

John Mack

37~

44

48

The Japariese Abacus

What are the odds that (~) is even?

Perdi~<

* * • * *

Marta Sved

54

55

61

THE FRONT COVER

The British epidemiologist and pioneer of tropicalmedicine, Sir Ronald Ross (1857-1932)1 is bestremembered today for his life's work on the transmissionof malaria, for which he was awarded the Nobel Prize inPhysiology and Medicine in 19020 It was Ross whoshowed conclusively that malaria is transmitted bymosquitoes and who analysed that transmission with aview to control or eradication of this virulent diseaseo

Ross was a considerable mathematician as well asbeing a medical re~earcher (he also wrote q,-:!itemer~torious poetry)w Among the studies he made was thedevelopment o~ a mathem~tical model of malaria and itstransmission 0 This led to what are now called the~Ross

Halaria Equations. Our front cover $hows the so-calledtrajectories of these equationso

Let a population of mosquitoes live in the samelocality as a population of humans and let x be' theproportion of mosquitoes infected with malaria and ythe population of humans so infected. In given time~ aproportion R of the infected humans recover and aproportion H of the mosquitoes die of malariac Ifeach human suffers 8 bites in this given time and eachmosquito delivers b bites in this same time~ then inthe long run

Bb-RHx x B(!f+b) ( 1 )

Bb-RI1y = y

b(B+R)

provided these quantities ~re positive.

If x ~ ;~ y ~ y , then we may plot x,ygraph as shown in Figure 1. _Th~ values of x~y

the arrows until th~ values x~ yare achieved.that for realism we assume

on afellow

Note

o < x <' 1 o <: y -<

The diagra~ shows the case for which x = 0.35 yOw30 ..

If x < 0 ~ or y < 0 ~ then the situation xx, y = y cannot be achieved and the arrows show that inthe long run x = O~ Y = 0 0 This cor~esponds to therebeing no malaria present - a much wished-for situatione

35

Figl1.re. 1 ~

Equations (1) show us that if Bb > RH , then bot~

x ~ yare positive and the situation'is that of. Figure1 which corresponds toa·situation of endemic malariaoIf, on the other hand~ Bb < RM , then both x~ ~ arenegative and so the malaria is ~radicated.

In order to eradicate malaria therefore we need toachieve a situation in which

RH > Bb (2)

That ~s to say we need to combine several factors:

Increase the recovery rate of the humans,

Increase the mortality of the mosquitoes,

Decrease the rate at which mosquitoes bite humans.

Inequality (2) shows an impo~tant point. It isnot necessary to eliminate all mosquitoes~ nor to cureevery human case of malaria, nor need all biting beprevented. As long as Inequality (2) continues toapply, malaria cannot spread in .the population. Thisled to considerable hopes that malaria might beerad1cated~ as smallpox has been.

36

Unfortun~tely, matters not taken into account 1nthe analYsis complicate mattersz Malaria olasmadia ar?becoming resistant to the drugs (derivatives of quinine)used to keep the value of K high~ and mosquitoes havebecome resistant to the pesticides used to keep ~ne

value of H upc It has also been discovered thatmalaria. infects other animals besides humans and 1:n~s

corresponds (in essence) to higher values of 0 andlower values of R than we might at first supposez

Nonetheless the analysis has been very helpful~ anacontinues to be helpful~ in malaria control programs.

In 19~3~ the American Journal of Hygiene devoted a,special supplement to the Ross Malaria Equations. Mostof this was written by Alfred J. Latka.!, ,-3 mathematicianat the Johns Hopk~ns University iR Baltimore. (We tend'to fo~get that malaria was endemic in Baltimore and inmany other parts of the United S~ates until well intothis centuryv) Latka produced a yery full study of theequations in the course of this work~ and Figure 1 aswell as the simplified form of this, that appea~s on thecover)· is based on one of his diagramsc

* * * * *He studied and nearly mastered the six books of

Euclid since he wa~ a member of CDng~essc

He began a course of rigid mental discipline withthe intent t~ improve his faculties~ e~oeciallv hispowers of logic and language. Hence his fondness forEuclid~ which he carried with him on the ·~i~cuit till hecould demonstrate with ease all the proDositions i~ thesix books; often studying far into the night, with ~

candle near his pillow,. while his fallow-lawyers, hal adozen in a rQom~ filled the air with intermina Iesnoring. - Abraham Lincoln (Short Autobiography~ 1860

* * * * *

THE $QUARE ROOT OF

37

ZERO

Michael A~B~ Deakin} Monash University.

If we confine ourselves to the study of realnumbers~ then it is impossible to ascribe a square ~oot

to a neaative number. In particular~ the number -1 hasno squa~e root.. But we"have come to" accept that we canhave another~ richer number system, that of theso-called complex numbers~ in which -1 has not just onebut two square, roots

(We need to have two square root~ to achieve consistencywith real arithmetic, e.gu (± '2)~ = 4~ etc.)

Complex numbers are the numbers of the form a + ib~

where a and b are real numbers 0

At first~ complex numbers were r~garded with somesuspicion, but in due course they b~came accepted.

Complex algebra leads to very elegant theorems andresults and in some ways is simpler than real algebraaAll the familiar rules apply~ including the following:

If c1

and

c 1c 2 = 0 ~

c~ are complex numbers and~

then eith~r c1

= 0 or

In complex algebra, every quadratic equation hasexac~ly two roots (although in some cases these may beequal>, every cubic equation has exactly three roots(again with possible equalities), every quartic equationhas .00' and so on. These properties are riot shared byreal numbers.

In 1873, the BrLtish mathematician Cliffordconsidered a different algebra, whose elements were o~

the form a +' &b where a, b are real numbers and £4

= 00 [& is the Greek letter epsilon.] True, theredoes not seem to be the same need for these as there is~or the complex numbers, as 0 already has a perfectlygood square root : itself.

However these new numbers (they are called dualnu.bers) have some applications in mechanics (which iswhat led Clifford to consider them) and have found otheruses since.

First let us see how they work. Let

38

be two dual numberso Then it is required that c 1when and only when a

1a

2and b

1= b

2Also

can add them

we

( 1 )

to form a third dual number~ or we could subtract them

(2 )

T~en we might multiply them

since 2& = 0 ..

(3)

Addition and multiplication obey all the usual rules ofalgebra bar one: the product of non-zero dual numberscan be zeroo

[It might at first sight seem that this -equationtells us tha~ £ = O~ but it does not~ It is &~ thatis zerou & b

1and & b

2are two dual numbers~

distinct <unless b1

or b2

is·zero) from zero .. ]

Let us investigate the division of dual numbers 0

Put

and seek to find a3~ b 3 ..

From the definition of a quotient, we must have

Then

unless

So

(> We also have

39

again unless at = o.

What we have just done is to show that, except inthe case of at = 0, ~e have

+ e (4 )

A shorter way of 'working out the result of dividingc 2 by c

1' that you might prefer, is as follows:

C2 a"", + & b"", (a.., + £ b 2 ) (a 1 - 1£ b

1)

JIt. JJt:. A-

c1

a1

+ & b1 (·a 1 + & b

1)(a

1& b 1.)

G2

a1

+ & (a1

b2

- a2

b1

) _.£2b2b1

2 ?a

1- (£ b'

1) -

a2

a1

e(a1

b2 - ~2bi )

2+

a~2

a1

since & = 0,

which reduces to (4) •

Formula (4) does not work in the case a 1 0

We may not divide by any number of the so-called puredual form e b ~ just as in real algebra we may notdivide by zerOD

- 1. __- ....._ ......~"""1.L +- ... ~+- rflf.::i-.l

40

algebra is consistent is to note acorrespondence between the dual number2 x 2 matrix

(~

straightforwarda + & b and the

(5)

It is possible~ and indeed . not difficult~ to gothrough the algebra of such matric~s and check all theformulae given above. For example & corresponds to

(~ b)and this matrix~ multiplied by itself, gives

which corresponds to O. This gives a clear ~ystem inwhich &,0 'are manifestly different.

It is easy to see how. Equations (1)~ (2), (3)translate into this new ma~rix notation. To checkEquation (4)~ note that as we go over to matrices

corresponds to [

a"""-'o

(or we could write this matrix product the other wayaround) •

Now, remembering that a1

~ 0 !I

(:1 b 1] -1 1 (:1 - b 1]at 2 a

1al

and so

(~2 b2 ] [:1 b 1] -1 1 r2b

2] r1-b 1]

a2

a1

2 o a.., o atat ....

1 (afl a 1b 2 -a2 b 1]

2" a2

a1

and this corresponds to Equation (4).

One use for dual numbers is in the analysis ofsmall errors. Suppose, for instance, a rectangle haslength a~ and width a_ and that the lenath a. has

41

associated with it a (relatively) small error b1

That is to' say, if L is the true length!! then

a1 - b 1 < L < at + b

1(6 )

·Similarly~ there is a relatively small error b2

~ 0 in

the measured width a~~

('7 )

The true area LU thus satisfies

Io@o a1

G2

- (a2

b2

+ a2

b 1 ) + b1

b2

S LU < a t a2

+ (d1

b2

+ a2

d1

)+ b1

b2

Remember now th~t b1

, b2

are relatively small, and so

b1

b2

is very small when compared to a 1a 2 For

many practical purposes we may simply neglect 'the b1

b2

and write

(8)

If ~e now c6mpare Inequality (8) with Inequalitiesand (7) arid the way these were set~p~ we see thatcan say that (to a good approximation) the area is

with a relatively small error of ~lb2 + a2 b 1 •

This, may be done very efficiently using dualalgebr~D Write the lengt~ as at + & b 1 and the width

as a 2 + £ b 2 - Then the area is (al~£ bt

)(a2 + & b t )

and we can multiply this out to give a1

a2

+ & (a1

b2

+

a2

b1

>, obtaining the estimate of the area and its

associated error both in the one calculation.

Dual algebra was initially proposed by Clifford asa means of dealing with certain problems in mechanicsand in geometry. This work was taken up by Study inGermany (1903)and by'Kotel'nikov in Russia (1895-1899).The best English summary of this work is to be found inChapters 2, 3 of Brand's Vector and Tensor Analysis <NewYork : 1947) but it is ~ather ~oo technical to go into

~ here. In particular, it considers the situation wherethe a~b occurring above are not numbers but vectors.<Cli~~ord~s original study began, in point of fact~ withan even more complicated case.)

42

Another ~ecent applic~tion of dual algebra concernsthe foundations of calculus. Newton's original versionof ~his made use of so-called infinitesimal quantities:numbers that~ seen from som~ points of view were zero~

but from other points of view (for example when wewished to divide by them) were not. This ratherunsatisfactory situation was brought toa focus by thephilosopher (and bishop) Berkeley in his book TheAnal y s't (1.734).

Berkeley· accepted the utility of calculus whilepointing to logical inadequ~cies in {ts foundations.This unsatisfactory state of affairs remained until thework of Cauchy (1789-1857) who succeeded in prOViding alogically watertight basis for the calculus, by avoidingentirely all reference to infinitesimals. This hasremained the standard approach to calculus to this day.

Nevertheless~ as Abraham Robinson wrote in his bookHon-Standard Analysis (Amsterdam: 1966)!' "in spite ofthis shattering rebuttal [Bishop Berkeley's argument],the idea of infinitely small or infinitesi.al quantitiesseems to appeal naturally to our intuition .. " Robinsonset himself the task of reinstating the infinitesimal~

and he succeeded in doing so in a series of researchesundertaken in the 1950 1

50

Beginning with an idea first mooted, but notelaborated~ by· Newton's great contemporary Leibniz, hesucceeded in producing an ~lgebra that included ordinary(real and· complex) numbers and infinitesimal quantitiesin one unified whole.. (Much as the system 0+ comple~·~

numbers includes real and imaginary numbers in adi+ferent such whalea> The approach is now termedno~-standa,d analysis and i~ provides an alternative(but equally rigourous} approach to the foundations ofcalculuse

Robinsonls account is technical 'and difficult!!although it succeeds in its aim of reinstating theinfinitesimal as a valid concepto More recently (1982)Dawn Fisher suggested the use of dual algebrao In herversion infinitesimals are numbers of .the form e b!l iDe ..neither real numbers nor zero", ··Whether this much,simpler version can do all that Robinson's can is. notentirely clear-a This is a matter time may perhapsclear up.

Interestingly, although in his initial paperClifford did treat ~ual algebra incorporating theelement &~ for which &~ = O~ he spends much more time ona related system. This is the so-called duo orduplex ~lgebra involving numbers of the farm a + w b ,where w4 = 1 0 [w is the Greek letter omega.]Again, require that a 1 + w b

1= a 2 + wb

2only if a 1

a 2 and b1

= b2

- Dupiex algeb~a.proceeds very like

of Equations (1), (2) need not detain us at all,the analogue of Equation (3) is the slightlycomplicated

43

whilemore

(a1

a2

+ b1

b2

) +w(a1

b2

+ a2

b1

)

(9 )

To divide duplex numbers!' proceed as we did insetting up Equation (4). . The result is (multiply topand bottom of the fraction on the left by a 1 - w b 1 )

a~ + w b", a2

G1 - b

2b

1a

1b

2- a

2b

1.4. .:. (10'>= 2 .., + w2 b 2 .at + w b

1 b"a1 1

a1 1

This result holds as long as b1~ ± ~10

in other words, divide by multiples of 1product of two non-zero duplex numbersThis is seen in the equation

(1 + w)(l - w) = 0 •

We may not!'

± w Thecan be zero.

( 11 )

Duplex numbers are less used nowadays than dualnumbers, but they find some applications in the theoryof relativity and they have a ready interpretation inordinary algebra a + w b corresponding to a ± b F a- w b to' a + b 0

Finally we may mention that bothcomplex numbers have matrix interpretationsthat of Matrix (5). We have thecorrespondences

duplex andsimilar to

following

a + i b to (a :)-b

a +& b to ( a :)0

~"a + w b to ( a :)b

(Can you see the pattern underlying these matrices?)",-Ir

44

CLRAR THTNKTNG ABOl_TT LTMTTS

Alan Pryde 7 Monash University

Mathematics, ever so slightly abused, can be usedto obtain some interesting conclusionso Consider forexample the following prriblem: given a non-negativenumber L find all solutions x > 0 of the equation

~(x + y(x + y(x + ORO) L ( i )

It is probably fairly clear what the left side ofequation (1) means - tryere is an infinite number ofsquare roots to be taken~ with x added each time butto be precise we may proceed as follows. ~et a

1= ~x~

=2 = -Icx + a l ) , Ci3 = -{(x_+ «3 2 ) DOO!' an Y(x + an-i)

~or n > 1 a Then the problem is to find x such"that

lim a = L It

n(2)

Some might argue, however, that we can solve (1)without mentioning limitsu Indeed, squaring (1) we

obtain x + L = L2 so that the solution is x = L

2L.

matter: theexperimenting

we find this

This seems to be the end of theproblem is apparently solved. However~

wi th d"if-ferent choices of the number Lis not the case ..

L = 3: Then x "(2 - L = 6 and equation (1) becomes

3 • (3)

L 1 and equation (1) becomesThen x

approaching

= 2cr984eeo, a 2 = 2.907 GOS' a 3a sequence apparently

2 .. 449Indeed at

ii 4 = 2 .. 997

L 3.

1(1 + YS>2

(4)

which is the answer to problem 10.3.2 of this magazine.

L = 1 Then x = 0 and equation (1) becomes

-{( 0 + y( 0 + ~(O .... 0 ) = 1 .. (5)

L o Again '-x o , though equation (1) becomes

45

~(O + y(O + Y(O + 000) = 0 0 (6)

Now (5) and (6) cannot both be true and ~o we arefaced with a. dilemma. Clearly <6> is correct and not

(S). Therefore our solution x = L2~ L is .not validfor all choices of L > 0, and there must be doubt asto its validity for any particular L". For example,can we be sure that (3) and (4) are correct? On theother hand, the computations that led to the solution

x = L 2 - L were valid, weren't they?

One of the' features of modern mathematics is itsability to resolve such dilemmas. The problem with our

solution x = L2 - L of equation (1) is that we have madea hidden assumption which is not always a valip onewe have assumed that equation (1) does have a solutionfor any given L ~ 0 In other words, what we have

done is to show that "if there is a solution of (1) then

it has to be x = [2_ L ~ In reality therefore, we haveonly half solved our problem. The more difficult halfis to determine "the values of L for which there is asolution. To do this require~ some interesting,perhaps difficult, mathematics~ and the final answer is~

I think, rather surprising.

Our a±m is to de~ermine thoseequation (1) is satisfied for some x ~

by proving

Lo .

for whichWe do this

for each x ~ 0 , L = lim a exists.n

n-+oo(7)

Clearly"~hen L exists it is ~ 0 0

Having done this, our previous computations show?

that if L denotes the limit then x L- L or

L2 - L - x = 0 . So L !( 1 ± y( 1 + 4x) ) and we must

2

also recall that L ~ 0 . If .)( = <) then L = 1 or 0;

but as. we have already observed, (5) is incorrect, andso L = <) If x ::; 0 !II since L ~ 1 , then

L = !( 1 + l'< 1 + 4x) ) which!, as x yaries, assumes all2

values L > 1 ·50 equation ( 1 ) is solvable for L 0

and for each L > 1 . In each case the solution is?

-)( = L- - I

46

Our final task is to prove (7).

A very useful theorem says that if {an} is a

sequence which is both increasing (i.e. an ~ a n + 1 for

all n) and bounded above (i.e. there is a number Ksuch that a < K for all n) then lim a exists.

n = n

We now show that our sequence an has these two

properties~ i.e. that it is increasing and boundedabove. .

First we show it is increasing. Certainly sincex ~ 0 a l = ~x ~ 0 a Then a 1 = ~x ~ y(x + a 1 )

since a1

~ 0 in turn it then follows that

=2 = -{(x + a l ) ~ ~(x +.Ci2 > = Cil 3 since at ~ Ci2 ; and so

on: once we have shown that ak~l ~ a k ~ for any k~ we

can then use this to show that a k = ~(x + ak

-1

)

~ y(x + ak

+1

). This procedure shows that, for all n,

an ~ an+i· Thus our se~ue~ce is increasing.

To show that lim a exists it remains to shown

n ...oo

that the sequence a is bounded abovelll We show thatn

!( 1in fact, for a.ll n ~ a ~ + y< 1 + 4x». We shown 2

this step by step just as we showed that the sequence

an was increasingu Certainly~ for n = 1

at = ~x = ~Y(4X) ( ~(1 + Y(l + 4x»; then~ using this

result~ for n = 2 ,we have

~(* + ~ ~(1 + 4x) + *<1 + 4x»

1~ ~{1 + 2Y(1 + 4x) + {~(1

1 ~(1 + ~(1 + 4x»22

1 (1 + .y{ 1 + 4x» D

La

thatyou

.3,.....L

Now if you look at the aGQve calculationthe only assumption used in showing that1 1

~ 2 (1 + ~(1 + 4x) is that a1

~ 2 (1 + ~(1 +

47-

see

1

Hence~ using a 2 ~ ~(1 + Y(i + 4.~·) it immediately

follows that ~(v + a_' i k~l + ~(1 + ~X}); andL -.L.

sa

on1

we next get d4

~ ~(1 + ~(1 + 4x)) .3.na similar-I.v

then for ar:=:' a _So~ ff::r a.l ":! r •6 !l ...

1.-J

1 --I ( 1 4."..-) '\.a ~ ~( + + I ..n .L

Thus ~(1 + Y{l + 4x)..:-

is an upper bound fer the

sequence an

So, by the theorem we quoted~ lima "00n

a n

It is by no means uncommon f9r mathematicians~ ~r

those using mathematics~ to approach a problem with theunspoken aS5umotion that there is a solution to a givenequation. This is a small abus~ of mathematics whichfrequently will not produce any -difficulties.. However =

as our example above should demonstrate~ the techniqueis not infallible: To overcome the adverseconsequences~ like eauation (5)~ deeper arguments andmore interesting, though perhaps ha~der~ mathematicsma.y be required ..

Bertrand Russell

HMathematics may be defined aswe never know what we are talkingNhat we are saying is true.!1

the subjectabout" nor

in ~..,hich

~..,hether

-4\ ..

"Pure mathematics consists entirely of assertion:­to the effect that~ if such and s~ch a oropcsitian istrue of anything~ then such ~nd such another propositionis true of that thing. !~ is essential not to discusswhether the first oroposition is really true~ and not tomention what the anything is~ of which it lS suoposed tobe true.1!

SIMPLE RATIONAL NUMBERS

John Mack, University of Sydney

In these days 04 calculators and computers, decimalnotation reigns supreme upon the .screen and moreover,what we see is usually truncated or rounded to say 8~ 10or 12 decimal' places unless. we are using an exact,multiprecision .or infinite precision arithmetic packagefor ~ourcalculations•. For 's~all' integers, assumingour calculating device is decently. accurate, (as most.are), then we expect the results of arithmeticoperations to be exact until we exceed. the display size.It then becomes an interesting problem (and one ofextreme importance) to decide whether or not one can useone's calculating device to obtain exact answers toproblems in integer arithmetice

For example, ~f you have a standard scientificcalculator, it is fairly easy to discover the highestpower of 2 where size will fit exactly into your screen

30display. If it is, say, 2 , then the challenge is towork out how, using your calculator, plus pencil and

paper~ you might find the exact value of 2 40 withoutdoing 'unnecessary' work.

ais0.125Whileis not always finite.1

The problem we are going to discuss here is .of adifferent kind and concerns the fact that the decimalrepresentation o~ rational numbers lying between 0 and

18

1terminating expansion, 3 0.333 ••• is noto It "is

fairly easy to prove directly that if x lies betweeno and 1 and has a terminating decimal expansion~ thenx is a rational number which can be written in the formp/q~ where the ~nly possible prime divisors of q are 2or S.

With a little more work~ it is possible to show·that the decimal expansion of a reduced rationalfraction x = p/q will always terminate (remember' thata fraction p/q is. said to be reduced if p and q.have no common factors other than ± 1), unless someprime other than 2 or 5 is a divisor of q .and whenthis occurs then the decimal expansion of x isinfinite and 'eventually periodic'. This means that~

possibly after an initial block of digits, the rest ofthe expansion consists of a certain finite sequence ofdigits which repeats over and over again. For example~

1

15= 0.0666 ••• D

49

has an initial block consisting of the single digit 0followed by the digit 6 which repeats, while

5

44= 0.11363636

has an initial block of lengthrecurring sequence 36 and

2 ~011owed by the

1

41= 0.02439024 DDllI

has no initial block, just the recurring sequence02439.

Some experimentat10n with a calculatoryou' to guess when there is and when thereinitial block.

will help­is not an

The upshot of all this is that if we are relying ona limited display, decimal representation calculator,then we will often be unable to tell, from what we see,whether or not we actually have a terminating oreventually periodic decimal representation and so unableto tell whether or not we may be dealing with a rationalnumber. This leads to the question "Given a truncateddecimal, can we guess whether or not there is a 'simple'rational number that might be close to or equal to it?"By 'simple', we mean ~ne with a denominator smallcompared with the length of the decimal, for, after all,every finite decimal isa ration~l nt.J",ber.

For example, suppose we read 'O.02439~ on ourdisplay. This is the rational number 2439/10000, butour question is "ls there a rational number p/q !'

with q much less than 10000, which could b~ the exactanswer to our problem?" From the example above, we see

1that, indeed, might well be the actual answer and we41

could re-examine the steps which led to our calculated

answer to see whether 1 is in ~act correct.41

There is a technique ~or finding such simplerational numbers. This technique~ called the continuedfraction method, produces a sequence of rationals withincreasing denominators that lie close to a givennumber. Rather than discuss the technique in generalte'rms, we shall apply it' to a specific problem.Afterwards, we'll give some references' which contain afuller account of the method.

50

Here is the problem. . You are at a meeting atwhich someone~ in order to support an argument, says "lhave donea~urvey on this issue and found that 58~6% ofthose surveyed said ,'Yes', while oAly 31X said 'No' ~ ~o

clearly there is str9ng support for my proposalm" Theproblem is that the number of persons surveyed is notstated and you would like to know hciw s.al1 a samplecould be surveyed to produce results as quoted. 58.6%

586 293 -1~ 31 155 hmeans or ~ h means , or ___, so per aps1000' 500 100 500

the survey. really covered 500 people~ which is animpressive number. But wait a minute. Quoting to 1decimal place suggests rounding off and so we arelooking for percentages in the range 58.55 to 58.64 and30.95 to 31.04, assuming the speaker is giving accurateestimates. Perhaps .there .is a denominator q (equal tothe survey size) 'which will produce percentages 100P1/q

and 100 P21q within the above ranges~ with q much

less than 5007

To apply the ,continued fraction method, we need tounderstand how' it workso Given any number x betweeno and 1 , it will give us a sequence Pl/ql~ P2 /Q2'o.o

of rationals~ alternately larger than and smaller thanx which close down onto x:

xt , I I I

p~ P4 P3 P14

~ ~

q2 q4 q~ ql~

Taking x 58.55% 0.5855 = 1171/2000, we obtainits continued fraction as follows:

1171 1 1x =

2000 20001 +

829

1171 1171

1

1 + 1

1171

829

1

1 +1

+342

829

51

1

1 +1

829

342

+

1 +

2 + 145

3421 1

1 11 +

1 1 +1

1 + 1 1 +1

2 + 342

145

2 +

2 + 145

1

1 +

1" +1

1 12 + 1

2 + 145

2 +

2 + 41

52 2 + 52

1 1

1

1

2 +

2 +

2 +

1 +

1

1

2 + 1

2 + 52

2 +

1 +

1 +

41 1 + 41

52

(after several more such steps)1

11 +

1 +1

2 +1

2 +1

2 +1

11 +

13 +

11 +

12 +

11 +

2

The numbers Pnlqn are found by calculating the values

of the intermediate steps:

Pi 1 1 P2 1 1 P3 1

"q 1 1 1~ q2 ! ""'!Iq3 1 '"1 + 1 + "11

1 + r"\Ji!.

P4 7 Ps 17 P6 24 P7 89 Pa'q - ,

q4 12 qs 29 41 q7 152 qa6

3

2 5!i

1 + 3'

113

193

315

538 731 ' ql1

1171

2000 (as it should be!).

The fractions greater than 0.5855, with denomi"nators1 3 17 89

less than 500, are 1 ' 5 ' 29 and 152 • Of . these,

17 189the ones less than 0.5864 are 29 and

152 •

We now see if there is a fraction p"",/29 lying in.:.

the range 0.3095 to 0.3104. Since.3we try ·9/29 and find 9/29 = 0.3103range!

of 29 is about 9~

is within the

53

Thus our speaker could have surveyed as few as 29people and obtained the quoted percentage responses.29 is not nearly impr:-essive a size as 500 and may wellbe an unimpressive size in terms of the argument thespeaker is supporting~

"If you have followed the above calcu~ations, thenyou would have despaired over calculating the Pnlqn by

successively simplifying the fra~tionsg Fortunatelythere is an easy way to do it : if we "have

1x = 1

1

1a~ +

...:-

Ci 4" +

Pi 1 P2 1a 2

first calculate and1 + 1ql a

i q2 /" a1

a2

iii +i3

2

Then~ miracle of miracles, we can compute

n ~ 3 by the recursive formulae

for

2

Check it out on the above example! If this leadsyou to believe that it might be true, you can eitherprove it or find it in an account of .continuedfractionSa "This may be" found, fo~ example, inHDDavenpo~t's lovely little book The Higher Arith.eti~

published by Hutchinson, (reprinted by Harper andBrothers, N~w York, 1960) , which ought to be in everyschool library!

Another account can be fGund in Volume II ofC=VcDurell and ADRobson's Advanced Algebra~ a book that40 years ago used to .be' a standard H 0 s. C a textbook, "andso is probably in many ~chool libraries. Cbntinuedfractions are first introduced on po242o

FinallY9 a note of thanks from the author toDr Henry PQllack~ who recently retired from BellResearch Laboratories in the U.SaAo and who mentionedthis niceappl~cation of continued fractions to guessingpossible sample sizesc

54

THE JAPANESE ABACUS

An exciting contest between the Japanese abacus andthe electric calculating machine' was held in Tokyo onNovember 12, 1946, under the sponsorship of the U.S.Army newspaper, the Sta'rs and Stripes. The abacusvictory was decisive.

The Nippon Times reported the contest as follows:"Civilisation, on the threshold of the atomic age~tottered Monday afternoon as the 2,OOO-year-old abacusbeat the electric' calculating machine by adding,subtracting, dividing and·a problem including all threewith multiplication thrown in, according to UnitedPress. Only in multiplication alone did the machinetriumph. II

The American rgpresentative of the calculatingmachine was Pvt. Thomas Nathan Wood of the 240th FinanceDisbursi~g Section of General MacArthur1s headquarters,who had been selected in an arithmetic contest as themost expert operator of the electric calculator inJapan.The Japanese representative was Mr KiyoshiMatsuzaki, a champion operator of the abacus in theSavings Bureau of' the Ministry of Postal Administrationw

The abacus scored a total of 4 points as against 1point for the electric calculato~. Such results shouldconvince even the most skeptical that, at least so faras addition and subtraction are concerned~ the abacuspossesses an indisputable advantage over the calculatingmachine. Its advantages in the fields ofmultiplication and division, however~ were not sodecisively demonstrated.

The Abacus Committee of. the Japan Chamber ofCommerce and In~ustry says that,- in a contest inaddition and SUbtraction, a first grade abacus operatorcan easily defeat the best operator of an ~lectricmachine., solving problems twice as fast as the latter.In multiplication' and division the margin of advantageover the electric calcul~tor disappears when there aremore than a total of 10 digits involved.

From The Japanese Abacus~ by V.Yamazaki, Prentice-Hall,1965a

* * * * *

55

"V\lHAT

THAT

ARR THE ODDS

f{ n) .TS RVEN?..... r '

Marta Sved~ University of Adelaide

(Dr]You are probably familiar with the expression

(read: n choose r), which is the number of ways inwhich a subcommittee of r members can be chosen out ofa committee consisting of n individ4alsc The symbol

~) is an abbreviation:

~) =1)(1'1 - 1) .. III (n - r + 1)

1 .. 200 r

giving the required numberso

The numbers(

Tly

] play a very distinguished role in

algebraic expressionso The best known of theseexpressions is the binomial expansion:

(a+b)1) = an + + .( n] n-l+ 0 .. + - ab +b #

n-1

To make this expression" more uniform looking, we define

(~) = (:) = 1 •

which makes sense when we consider that there is exactlyone committee with no members and one way of choos~ng

everybody for a subcommittee 0

obtain

n- O1"1 1~ 2n 3n 4Ti 5n 6n 7n 8n 91"1 10

11

89 1

45 10 1

11

621 7

2884 36

120

3570 56

126252 210

21 3528 56

, 84 126120 210

11 1

2 11 3 3 '1

1 4 6 41 5 10 {O 5

1 6 15 20 151 7

1 81 9 36

1 10 45

When this is done then the binomial theorem reads

(a + b)n = (~)an + (~J;!n-lb + ••• + (n~l)abn-l + (~)bTl.

You are also l~kely to be familiar with the Pascalarray which you obtain when writing down in successionthe binomial coef+icients occurring in the expansions O?

012(at + bJ , (a + b) , (Ii + b) and so on Cl You

56

This array provides endless fun, if you like tw playwith numbers. One of the most important oroperties ofthe array is expressed by Pascal's Recursion ~ormu~a:

( TI + 1) = (Tf') + (: )r+l r r.t.

You can check this~ (proving is not difficulteither)~ lookina at the array and seeing how the entriesin each row can be obtained from the entries of theprevious row by just adding each time two neighbours.In fact~ this is the way of getting quickly thecoefficients in the expansions of "(a+b)n withincreasing T1

The title suggests that we want_to deal with" theevenness or oddness 0+ these binomial coefficients. Aglance at the array already suggest·s tha.t there is anabundance of even numbers in the Pa~cal array. Lookingfor e}~ample at rows n = 2 :r n = 4,- Tl = 8, you ·find thatexcepting the first and last numbers in the raw~ alwaysequal to 1, all the other entries are even. Howeve~~when you look at the rows above these rows, that is atrows n = 1~ n = 3~ n = 7~" then you see that all thecoefficients are odd <without exception).

There is however a more efficient and also moreattractive way of settling the question of evenness andoddness. Instead of writing oui th~ binomialcoefficients~ we merely stat~~ whether they are even orodd, by writing 1 fer the odd coe~~icients and 0 far theeven ones. (Mathematically this means that we writedown the remainders after dividing each coefficient by2.) Next we note that we do not even have to loak atthe Pascal array tq obtain our new array~ because thePascal recursion formula gives us an easy rule to obtaineach row in succession~ since

even + even even wr t'e 0 + 0 0even + odd odd wr te (1 +odd + odd even wr te + 1 0

Thus we obtain the binomial array modulo -"'\ (see.,;:page 57)

57

58

If you have a large enough sheet of paper~ you mayco~~inue and enlarge the array. It will give you aneven better insight into the 'evenness-oddness problem,and if you paint your.O-6's red and your l-~'s blue, youcan use it as a wallpaper design or pattern for aknitted jersey.

You notice that to the array of ones and zeroes wealso add~d three different types of triangular frames.These may serve as further embellishments, but theirreal aim is to produce some mathematical results.

Triangles of type il which seem to repeatin a pattern are giv~n a thin frame. Next we note thatsuch triangles which we shall call cells seem to clusteron patterns of three:

These clusters will be framed by broken lines.Finally~ as far as our array goes, we note that theseclusters again cluster in threes~ giving a morecomplicated cluster which we mark with .heavy frames.You may guess that continuing the array you will findthat the heavy framed clusters will· make similarformations (you may use 'colours for the new~ largerframes) and so you may continue as far as time and papersheets permits.

What are the conclusions which can be drawn fromall this?

Let us first look at the reasons· for theseregularities. Note that there are unframed 'spacesfilled with zeros~ numbering 1 or 6 or 28 in our array 0

We shall call these spaces zero-holes. It follows fromour construction that the cluster arising bet~een theedge OT the Pascal array and a zero-hole .ust be thesa8e as the top-cluster oT the sa.e size arising ·betweentwo edges of the array with its vertex at the apex oTthe Pascal triangle. The firs~ ~ero-hole consisting of asingle zero entry is in the row n = 20 To its right

59

and left t~o cel~s appear identical to the cell abovethe zero entry. These.three cells together form ourfirst cluster. Looking a~ the last rows of thiscluster (replacing the row {I 3 3 1} of the Pascalarray), we see that all entries are equal to 1~ sincethis row is the amalgamation of the- last rows of twocells.

It follows again. from the rule: 1 + 1 = 0 that therow following the ~ast row of the cluster has only zeroentries between the two I-entries at the edges of thewhole array. This means that the entries of the Pascalarray which this row.replaces are even. (They are {46 4}. A larger zero ho~e is now generated consistingof three rows forming an inverted triangle. Itreplaces

of the Pascal arr~y.

4 610

20

410

Replicas of the cluster above the zero-hole appearon the sides and so we ,obtain the first heavy framedcluster, with its last row (n = 7) consisting of l~s

and followed by the next row (n = 8> where a'new, largerzero-hole, cons~sting of 7 rows originates. We note nowthat the first zero-hole begins at n 2~ the nextcentral zero-hole at n = 4~ and from the congruency ofclusters it follows that at n = 8, and generally at

n = 2k new central zero-holes arise, each larger than

the ~receding one.

binomial coefficients

We are able to state now that when

(nr

) are even.~ provided that

all

(> .:..... r ( n . ~eferring to the last rows of clusters~ wehave

if n = 2k - ~) is .odd for all r

These two last results can be proved by algebra, but itis interesting to see how they follow from the geometryof our .odulo 2 array.

Another consequence~ not so well known is that

_4th h ..c b . . (rn)n ( , ~ t e nu.ber 0, odd InoMial coeffIcients

Tor

is

Looking again at our picture~ we see that for n ~ 2there are 3 odd binomials~ namely

60

~) 1 (~) = G) = 1 •

Since the cluster ~or n < 4 contains three such cells 9

(nr

)there are 9 odd binomial coefficients when

Generally : the array for n < 2 kcontains 3clusters congruent to the cluster of entries for

n ( 2k - 1 , hence 3 times as many 1-s which signify oddentries in the Pascal array.

Thus : up to n = 8 ~ there arebinomial coefficients ~ up to n = 16(you can check this), and so. anD

3 x 9 = 27there are

odd81

We can conclude immediately that the probability ofa binomial being -even, approaches I (it is "almostcertain") as the Pascal array increases, although forarbitrarily large n we shall always have completely oddrows~ namely when n = 2

k- 1

even

array

:2+3+ •• +2

k+ 2N = 1

To be more precise, we find the numbers ofbinomial coefficients ~) when· n < 2

k•

The total numbers of entries in the Pascalfor n < 2 k is

since the number of odd binomials' is 3k~ i~ follows thatthe number o~ even coefficients is__ 22k-l

HE +

P(E)

Hence the probability that in the rangebinomial coefficient is even is

HE 3 k

H = 1- 22k-1 + 2 k - 1

considered a

Thus

P (E) ) 1 -,.,2k-l.4

=1-2~Y -"

where for large enough k , 2~)k can be made as small as

we wish. Thuswish by choosing

P(E) can be made as near tok appropriately large.

1 as we

61PRRDIX

The Australian Mathematical Olympiad for 1987 washeld on the 3rd and 4th o-f March" Seventy-one selectedcompetitors took part in the Olympiad, most chosenbecause of their performances in the A~5tralian

Mathemat~cs Competition. Some had ~hown their abilityin other competitions~ and a few had been accepted onthe st.rong recommendation o-f their mathematics 'teachers.'

For success in the Australian Olympiad, 11 goldcertificates, 15 silver certificates and 21 bronzecertificates were awarded. The remaining campetitorsall received participation certificates.

How would you have fared in this Olympiad? Thepapers se~'are appended. Have a go! Send me any o~

your solutions 9 or any queries you have about thequestionsQ

The Australian team ~Dr the Interna~ional

Mathematical Olympiad 9 this time being held in Havana 9

Cuba, is selected by taking the best six performers inthe Australian Mathematical Olympiad. This year·s team(in alphabetical order) is:

Chung Kim Van,

Jonathan Potts 9

Ben Robinson'9

Luke Seberry?

Terence Tao,

Ian .WanlessIj!

Duncraig Senior High School":Western Australia~

Brisbane Grammar School,Queensland!:»

Narrabundah College,Australian Capital Territory~

Sydney Grammar Schoal,New South Wales.

Blackwood HighSchool,South Australia.

Phillip College,Australian Capital Territory.

As usual a reserve member of theselected and he is:

team was also

Danny Culegar-i. Melbourne Grammar School"Victoria ..

There is still intensive training ahead for theteam. A highlight o~ this training will be a week inSydney at t:he IBM training school. At~ this school theteam, with its reserve~ will be joined by eight others~

selected also from their performances in the AustralianMathematical Olympiad 9 .and who also are in year 11 orbelow at school 9 so that they will still be eligible few·

62

selection for the 1988 Olympiad Team.

Those selected to joi~ the schooladdition to the team and its reserve are:

in Mav in

Geoffrey Bailey~

Robert Gates~

David Jackson~

Mark Kisi~~

Jeremy L~ew,

Martin O'Hely,

Blair Trewin~

Sam Yates~

St Aloysius College.New South Wales.

Knox Grammar SchQQ1~

New South Wales.

Sydney Grammar School~

.New South Wales.

Melbourne Grammar School~

Victoria.

Duncraig Senior High Schaol,Western Australia~

Salesian ColleQe~

Victoria.

Canberra Grammar School~

Australian Capital Territory.

St Peter's College~

South Australia.

The questions in the papers that follow prov~ae

difficulties of two kinds. The +irst kind ofdifficulty is that of understanding what the problem is.If you are not certain what the circumcircle of atriangle is then you cannot be sure you understandquestion 1. If you do not know -what a prime number is~

then you cannot do question 2.

The second kind of difficulty is that of solvingthe problem once it has been completely understood.

·Here there are at least two factors that come intp mastsolutions. The first is knowledge. This knowled~e

can be a knowledge of related mathematical results someof which might be necessary to use in solving theproblem. For example, in question 6~ you need to befam11~ar with proofs by 'mathematical induction to besure-~ooted in your solution. Familiarity with theproperties of similar triangles helps for question 1.

Let us assume you have all the necessary knowledge.It ~s then that the main <and intended) difficultyarises: conjure up sufficient ingenuity to out what youknow together to lead to a solution.

63

THE1981

AUSTRALIAN MATHEMATICAL OLYMPIAD

PAPER I

Tuesday, 3rd March, 1987

Time allowed: 4 hours

NO calculators are to be used.Each question is UJorth seven points..

Question 1

GKA is an isosceles triangle with base GKa! length 26.. GA and AK each ha.ve

length 4. Let C be the midpoint of AK and z the circumcircle of the triangle GCKo

Let Y be the point on the extension of AK such tha.t if E is the intersection of Y G

with z then EY is of length a/2.Prove that if z is the length of EC and y is the length of KY then ay = x 2 and

xb = y2 0

Question 2 .

Let p be a prime numbero Show tha.t the integer

(21') -2= 2p(2P-l)···(P+l)_2p p(p - 1)···1

is a multiple of p.

Question 3

In the country Patera there are 20 c;ities and two airline companies, Green Pla.nes and

Red Planes, to provide communication between the ·cities. 'J;he flights are atTanged

as follows:

(i) Given any two cities in Patera, one and only one of the companies provides direct

flights (in b~th directions and without stops) between the two cities..

(ii) There are two cities A and B in Patera. such tha.t a. journey cannot be made

from one to the other (with possible stops) using only Red Planes.

Show that, given any two citi~ in Patera, a. passenger-can travel from one to the

other using only Green Planes, making at most one stop in some third city.

h4

THE

1.987

AUSTRALIAN MATHEMATICAL OLYMPIAD

PAPER II

Wednesday, 4th March, 1987

Time allowed: 4 hOUTS

NO calculators are 'to be used.Each question is worth seven points.

Question 4

In the interior of the .triallgle ABC, points 0 and P are chosen such that. angles

ABO and CBP are equal, and angles BCO and ACP are also equaL

Prove tha.t angles CAO and BAP are equal.

QuestioI;l 5

Let m and n be (fixed) integers ~reater than 1, m even, and f a real-valued function,­

defined for all non-negative real nunlbers, that satisfies the following conditions:

(i) For all %1,%2, ••• ,x"',f{(:t'{' + %2' + ... + x:')/n) = ([f(Zl)]m + [f(X2)]m + ... + [f(xn))m)Jn;

(ii) 1(1986) =I- 1986;

(iii) [(198B) i: o.Prove that 1(1987) = 1~

Question 6

Prove that for each positive integer n (n > 1),

~. + v;. - v'2 > 1 + 1/v'2 + 1/v'3 + ... + l/v;..